The integrals of multivariable calculus

Multivariable calculus includes six different generalizations
of the familiar one-variable integral of a scalar-valued function
over an interval.
One can integrate functions over one-dimensional curves, two dimensional
planar regions and surfaces, as well as three-dimensional volumes.
When integrating over curves and surfaces, one can integral vector fields,
where the one integrates either the tangential (for curves) or the
normal (for surfaces) component of the vector field.

In this page, we outline the various integrals, methods you can use to solve
them, and their relationship to the fundamental theorems.

Line integral of scalar-valued function

The line integral
of a scalar-valued function $\dlsi(\vc{x})$ over a curve $\dlc$
is written as
\begin{align*}
\dslint.
\end{align*}
One physical interpretation of this line integral is that it gives the
mass of a wire from its density $\dlsi$.

The only way we've encountered to evaluate this integral is the direct
method. We must parametrize $\dlc$ by some function $\dllp(t)$, for $a
\le t \le b$. Then, we can calculate the line integral by turning it
into a regular one-variable integral of the form
\begin{align*}
\dslint = \dpslint.
\end{align*}

Note that the “length” $d\als$ became $\| \dllp'(t)\|dt$.
This quantity $\| \dllp'(t)\|$ measures how
$\dllp(t)$ stretches or shrinks the interval $[a,b]$ as it maps it
onto $\dlc$.

Line integral of a vector field

The line integral
of a vector field $\dlvf(\vc{x})$
over a curve $\dlc$ is written as
\begin{align*}
\dlint.
\end{align*}
One physical interpretation of this line integral is the calculation
of work when a particle moves along a path.
If $\dlvf$ were a force acting on a particle moving
along $\dlc$, then the integral would be the total work performed by the
force on the particle.

This integral is one of the most important of multivariable calculus.
We have
four alternatives to evaluate the integral, although most of the
alternatives work only in special cases.

We can compute the integral directly. We parametrize $\dlc$ by
some function $\dllp(t)$, for $a \le t \le b$. Then
\begin{align*}
\dlint = \dplint
\end{align*}

This method always applies. Sometimes, though, the integral will be
difficult or we won't even be able to evaluate it. Our lives can be
made easier by using one of the
fundamental theorems
to convert the line integral into something else.

Since this integral is really a
line integral of the scalar-valued function $\dlsi =
\dlvf \cdot \vc{T}$ where $\vc{T}$ is the unit tangent vector
\begin{align*}
\vc{T} = \frac{ \dllp'(t)}{\| \dllp'(t)\|},
\end{align*}
the formula for the direct method is the same as the formula for the
scalar-valued path integral with $\dlsi = \dlvf \cdot \vc{T}$.

We need to find a potential function $f$ so
that $\nabla f = \dlvf$. Then,
\begin{align*}
\dlint = f(\vc{q}) - f(\vc{p}),
\end{align*}
where $\vc{p}$ and $\vc{q}$ are the endpoints of $\dlc$.

If $\dlc$ also happens to be a closed curve, then the integral of
a conservative vector field $\dlvf$ will be zero.
Also, that if you know $\dlvf$ is conservative, another thing you can do is just change the curve
$\dlc$ to another curve that has the same endpoints as $\dlc$. In this
case, the line integral of $\dlvf$ over $\dlc$ is the same as the line
integral of $\dlvf$ over any other curve with the same endpoints.

If the vector field $\dlvf$ and the curve $\dlc$ happen to be in
two dimensions and if $\dlc$ happens to be a
simpleclosed
curve, then we
can use Green's theorem.
Green's theorem converts the line integral over $\dlc$ to a double
integral over the interior of $\dlc$, which we call $\dlr$,
\begin{align*}
\dlint = \iint_\dlr \left(\pdiff{\dlvfc_2}{x} -
\pdiff{\dlvfc_1}{y} \right) \, dA.
\end{align*}
Note that $\dlvf$ must be continuously differentiable everywhere in $\dlr$ for this to work.
Sometimes we write $\dlc= \partial \dlr$ to denote that $\dlc$ is the
boundary of $\dlr$. $\dlc$ must be oriented in a counterclockwise
fashion, otherwise, we'll be off by a minus sign.

If the vector field $\dlvf$ and the curve $\dlc$ happen to be in
three dimensions and if $\dlc$ happens to be a
simpleclosed
curve, then we
can use Stokes' theorem.
Stokes' theorem converts the line integral over $\dlc$ to a surface
integral over any surface $\dls$ for which $\dlc$ is a boundary,
\begin{align*}
\dlint = \sint{\dls}{\curl \dlvf},
\end{align*}
and is valid for any surface over which $\dlvf$ is continuously differentiable.
Sometimes we write $\dlc= \partial \dls$ to denote that $\dlc$ is the
boundary of $\dls$. $\dlc$ must be a positively oriented boundary of $\dls$, otherwise, we'll be off by a minus sign.

Surface integral of a scalar-valued function

The surface integral of a scalar-valued function
$\dlsi(\vc{x})$ over a surface $\dls$ is written as
\begin{align*}
\dssint
\end{align*}
One physical interpretation of the integral is the mass of a sheet with density given by $\dlsi$.

The only way we've encountered to evalute this integral is the direct
method. We must parametrize $\dls$ by some function $\dlsp(\spfv,\spsv)$, for $(\spfv,\spsv) \in \dlr$. Then,
\begin{align*}
\dssint = \dpssint
\end{align*}

Note that the “area” $d\sas$ became $\left\| \pdiff{\dlsp}{\spfv}
\times \pdiff{\dlsp}{\spsv}\right\| \,d\spfv\,d\spsv$.
The quantity $d\sas$ became $\left\| \pdiff{\dlsp}{\spfv}
\times \pdiff{\dlsp}{\spsv}\right\|$ measures how
$\dlsp(\spfv,\spsv)$ stretches or shrinks the region $\dlr$ as it maps it
onto $\dls$.

Surface integral of a vector field

The surface integral over surface $\dls$ of a vector field $\dlvf(\vc{x})$
is written as
\begin{align*}
\dsint.
\end{align*}
A physical interpretation is the flux of a fluid through $\dls$ whose
velocity is given by $\dlvf$.
For this reason, we sometimes
refer to the integral as a “flux integral.”

Like the line integral of a vector field,
this integral plays a big role in this course. We have three
alternatives to evaluate the integral, although most of the
alternatives work only in special cases.

We can compute the integral directly. We parametrize $\dls$ by
some function $\dlsp(\spfv,\spsv)$, for $(\spfv,\spsv) \in \dlr$. Then,
\begin{align*}
\dsint = \dpsint.
\end{align*}

This method always applies. Sometimes, though, the integral will be
difficult or we won't even be able to evaluate it. Our lives can be
made easier by using one of the
fundamental theorems
to convert the surface integral into something else.

Since this integral is really a surface
integral of the scalar-valued function $\dlsi = \dlvf \cdot
\vc{n}$ where $\vc{n}$ is the unit normal vector
\begin{align*}
\vc{n} = \frac{\displaystyle\pdiff{\dlsp}{\spfv}
\times \pdiff{\dlsp}{\spsv}}
{\displaystyle\left\|\pdiff{\dlsp}{\spfv}
\times \pdiff{\dlsp}{\spsv}\right\|},
\end{align*}
the formula for the direct method is the same as the formula for the
scalar-valued surface integral with $\dlsi = \dlvf \cdot
\vc{n}$.

If the vector field $\dlvf$ happens to be the curl of another
vector field $\vc{G}$, i.e., $\dlvf = \curl \vc{G}$, then we can
apply Stokes' theorem to convert the surface
integral of $\curl \vc{G}$ into the line integral of $\vc{G}$ around
the positively oriented boundary of $\dls$, which we
denote $\partial \dls$,
\begin{align*}
\dsint = \sint{\dls}{\curl \vc{G}}
=\lint{\dlc}{\vc{G}}
\end{align*}

Multivariable calculus student's don't typically learn methods to find $\vc{G}$ from
$\dlvf$. We can use Stokes' theorem to convert a surface integral
into a line integral only if we are told outright that $\dlvf =
\curl \vc{G}$ and are given what $\vc{G}$ is. But, if given the
surface integral that looks like $\sint{\dls}{\curl \vc{G}}$, we can
immediately recognize that Stokes' theorem is an option.

Stokes' theorem allows us to do one more thing to the
integral $\sint{\dls}{\curl \vc{G}}$. We can
switch the surface $\dls$ to any other surface $\dls'$ as long as the
boundaries of $\dls$ and $\dls'$ are the same, i.e., $\partial \dls =
\partial \dls'$ (assuming both boundaries are positively oriented). If $\dls$ is a complicated surface, we could feasibly save
ourselves some work by integrating over another surface $\dls'$ if that
surface is simpler than $\dls$.

If the surface $\dls$ happens to be a closed surface so that it
is the boundary of some solid $\dlv$, i.e., $\dls = \partial \dlv$,
then we can use the divergence theorem
to convert the surface integral into the triple
integral of $\div \dlvf$ over $\dlv$,
\begin{align*}
\dsint = \iiint_\dlv \div
\dlvf \, dV,
\end{align*}
where we orient $\dls$ so that it has an outward pointing normal
vector. This works only if $\dlvf$ is continuously differentiable
everywhere in the solid $\dlv$.

Double integrals

The double integral of a (scalar-valued) function $\dlsi(\vc{x})$ over a
two-dimensional region $\dlr$ is written as
\begin{align*}
\iint_\dlr \dlsi \, dA.
\end{align*}
One physical interpretation is the mass of a region with density \dlsi$.

We have encountered three alternatives to evaluate the integral.

We can compute the integral directly in terms of the original
variables $x$ and $y$. In this case, $dA = dx\,dy$.

We can compute the integral by changing to the variables $\cvarfv$ and
$\cvarsv$ by finding a function $(x,y) = \vc{T}(\cvarfv,\cvarsv)$. Then the integral
is
\begin{align*}
\iint_\dlr \dlsi\, dA = \left.\iint_{\dlr^{\textstyle *}}\right. \dlsi(\vc{T}(\cvarfv,\cvarsv))
\left| \det \jacm{\vc{T}}(\cvarfv,\cvarsv) \right| d\cvarfv\,d\cvarsv,
\end{align*}
where $\dlr$ is parametrized by $(x,y)=\vc{T}(\cvarfv,\cvarsv)$ for $(\cvarfv,\cvarsv)$ in
$\dlr^{\textstyle *}$. We sometimes write the determinant of the matrix of
partial derivatives
of $\vc{T}(\cvarfv,\cvarsv)$ as $\displaystyle \det \jacm{\vc{T}}(\cvarfv,\cvarsv) =
\pdiff{(x,y)}{(\cvarfv,\cvarsv)}$.

If $f$ happens to be equal to $\pdiff{\dlvfc_2}{x} -
\pdiff{\dlvfc_1}{y}$ for some vector field $\dlvf$, then we could use
Green's theorem
to convert the double integral
into the integral of $\dlvf$ around the boundary of $\dlr$, which we
denote $\partial \dlr$,
\begin{align*}
\iint_\dlr f \, dA = \iint_\dlr \left(\pdiff{\dlvfc_2}{x} -
\pdiff{\dlvfc_1}{y}\right) dA = \lint{\partial \dlr}{\dlvf}.
\end{align*}
To orient the boundary properly, outside boundaries must be
counterclockwise and inner boundaries must be clockwise.

We usually think of Green's theorem going the other way, i.e.,
converting a line integral into a double integral. One reason for
this is that multivariable calculus students don't usually learn methods to find $\dlvf$ from $f$.
We can use Green's theorem to convert a double integral into a line
integral only if we are told outright that $f =\pdiff{\dlvfc_2}{x} -
\pdiff{\dlvfc_1}{y}$ and are given what $\dlvf$ is. But, if given the
double integral that looks like $\iint_\dlr \left(\pdiff{\dlvfc_2}{x} -
\pdiff{\dlvfc_1}{y}\right) dA$, we can immediately recognize that Green's
theorem is an option.

As a special case, if we are given an
integral $\iint_\dlr dA$ (i.e., finding the area), we can let
$\dlvf(x,y) = (-y,x)/2$ so that $\pdiff{\dlvfc_2}{x} -
\pdiff{\dlvfc_1}{y} =1$ and $\iint_\dlr dA = \lint{\partial \dlr}{\dlvf}$.

Triple integrals

The triple integral of a (scalar-valued) function $\dlsi(\vc{x})$ over a
three-dimensional solid $\dlv$ is written as
\begin{align*}
\iiint_\dlv \dlsi \, dV.
\end{align*}
One physical interpretation is the mass of a solid with density given by $\dlsi$.

We have encountered three alternatives to evaluate the integral.

We can compute the integral directly in terms of the original
variables $x$, $y$, and $z$. In this case, $dV = dx\,dy\,dz$.

We can compute the integral by changing to the variables $\cvarfv$,
$\cvarsv$, and $\cvartv$ by finding a function $(x,y,z) = \vc{T}(\cvarfv,\cvarsv,\cvartv)$. Then
the integral is
\begin{align*}
\iiint_\dlv \dlsi\, dV = \left.\iiint_{\dlv^{\textstyle *}}\right. \dlsi(\vc{T}(\cvarfv,\cvarsv,\cvartv))
\left| \det \jacm{\vc{T}}(\cvarfv,\cvarsv,\cvartv) \right| d\cvarfv\,d\cvarsv\,d\cvartv,
\end{align*}
where $\dlv$ is parametrized by $(x,y,z)=\vc{T}(\cvarfv,\cvarsv,\cvartv)$ for $(\cvarfv,\cvarsv,\cvartv)$
in $\dlv^{\textstyle *}$. We sometimes write the determinant of the matrix
of partial derivatives of $\vc{T}(\cvarfv,\cvarsv,\cvartv)$ as $\displaystyle \det
\jacm{\vc{T}}(\cvarfv,\cvarsv,\cvartv) = \pdiff{(x,y,z)}{(\cvarfv,\cvarsv,\cvartv)}$.

If $\dlsi$ happens to be equal to $\div \dlvf$ for some vector
field $\dlvf$, then we could use the
divergence theorem
to convert the triple integral into the surface
integral of $\dlvf$ around the boundary of $\dlv$, which we denote
$\partial \dlv$,
\begin{align*}
\iiint_\dlv \dlsi\, dV = \iiint_\dlv \div \dlvf\, dV
= \sint{\partial \dlv}{\dlvf}.
\end{align*}

We usually think of the divergence theorem going the other way,
i.e., converting a surface integral into a triple integral. One
reason for this is that multivariable calculus student don't typically learn methods to find $\dlvf$
from $f$. We can use the divergence theorem to convert a triple
integral into a surface integral only if we are told outright that
$\dlsi = \div \dlvf$ and are given what $\dlvf$ is. But, if given a
triple integral that looks like $\iiint_\dlv \div\dlvf \, dV$,
we can immediately recognize that the divergence theorem is an
option.